Problem 193
Question
For the following exercises, find \(\frac{d^{2} y}{d x^{2}}\) for the given functions. $$y=x-\frac{1}{2} \sin x$$
Step-by-Step Solution
Verified Answer
\( \frac{d^{2}y}{dx^{2}} = \frac{1}{2} \sin x \)
1Step 1: Find the First Derivative
The function is given by \( y = x - \frac{1}{2} \sin x \). To find the first derivative, \( \frac{dy}{dx} \), differentiate each term.- The derivative of \( x \) is \( 1 \).- The derivative of \(-\frac{1}{2} \sin x \) is \(-\frac{1}{2} \cos x \) using the fact that the derivative of \( \sin x \) is \( \cos x \).Combining the two results, we have:\[ \frac{dy}{dx} = 1 - \frac{1}{2} \cos x. \]
2Step 2: Find the Second Derivative
Now, differentiate \( \frac{dy}{dx} = 1 - \frac{1}{2} \cos x \) to find the second derivative \( \frac{d^{2}y}{dx^{2}} \).- The derivative of \( 1 \) is \( 0 \), since it's a constant.- The derivative of \(-\frac{1}{2} \cos x \) is \( \frac{1}{2} \sin x \), as the derivative of \( \cos x \) is \(-\sin x \).Thus, the second derivative is:\[ \frac{d^{2}y}{dx^{2}} = \frac{1}{2} \sin x. \]
Key Concepts
Understanding DifferentiationExploring Trigonometric FunctionsSolving Calculus Problems
Understanding Differentiation
Differentiation is one of the fundamental concepts in calculus. It is the process of finding the derivative of a function. The derivative represents how a function changes as its input changes. If you think of a graph, the derivative at a point gives the slope of the tangent line at that point. It tells us how steep the graph is. By differentiating, we get a new function that shows this rate of change.
The first derivative is often denoted as \( \frac{dy}{dx} \) or \( f'(x) \), and it tells us about the slope at each point. In our exercise, we differentiated function \( y = x - \frac{1}{2} \sin x \) to get \( \frac{dy}{dx} = 1 - \frac{1}{2} \cos x \).
When we differentiate a second time, we find the second derivative, denoted as \( \frac{d^2y}{dx^2} \). This derivative tells us how the slope of the tangent line itself changes, giving us insights into the "curvature" of the graph, or how it bends. For our specific case, the second derivative becomes \( \frac{d^2y}{dx^2} = \frac{1}{2} \sin x \).
Understanding differentiation is crucial as it allows us to explore deeper into how functions behave and how we can predict these behaviors.
The first derivative is often denoted as \( \frac{dy}{dx} \) or \( f'(x) \), and it tells us about the slope at each point. In our exercise, we differentiated function \( y = x - \frac{1}{2} \sin x \) to get \( \frac{dy}{dx} = 1 - \frac{1}{2} \cos x \).
When we differentiate a second time, we find the second derivative, denoted as \( \frac{d^2y}{dx^2} \). This derivative tells us how the slope of the tangent line itself changes, giving us insights into the "curvature" of the graph, or how it bends. For our specific case, the second derivative becomes \( \frac{d^2y}{dx^2} = \frac{1}{2} \sin x \).
Understanding differentiation is crucial as it allows us to explore deeper into how functions behave and how we can predict these behaviors.
Exploring Trigonometric Functions
Trigonometric functions such as sine and cosine are essential in many fields, including physics, engineering, and mathematics. They are periodic functions, meaning they repeat their values in regular intervals. In calculus, understanding these functions and their derivatives is key.
For the function \( \sin x \), its derivative is \( \cos x \). Meanwhile, the derivative of \( \cos x \) is \(-\sin x \). These derivatives tell us how the sine and cosine functions change. In the context of our problem, we had to differentiate \( \sin x \), leading to \( -\cos x \), and then differentiate \( -\cos x \), leading to \( \sin x \). This cycle of derivatives is important to predict how trigonometric curves behave over time.
Trigonometric functions are always intertwined with their derivatives. Recognizing these relationships fosters a better understanding of complex calculus problems.
For the function \( \sin x \), its derivative is \( \cos x \). Meanwhile, the derivative of \( \cos x \) is \(-\sin x \). These derivatives tell us how the sine and cosine functions change. In the context of our problem, we had to differentiate \( \sin x \), leading to \( -\cos x \), and then differentiate \( -\cos x \), leading to \( \sin x \). This cycle of derivatives is important to predict how trigonometric curves behave over time.
Trigonometric functions are always intertwined with their derivatives. Recognizing these relationships fosters a better understanding of complex calculus problems.
Solving Calculus Problems
Solving calculus problems involves using different techniques to analyze and understand how mathematical functions behave. Problems often require performing a series of steps, such as finding derivatives, integrating functions, or solving limits. Each step builds upon the knowledge of fundamental concepts, such as differentiation and the behavior of trigonometric functions.
The exercise was a classic calculus problem where we were asked to find a second derivative. The steps took us from breaking down the given function into parts, calculating their derivatives, and applying these understandings to find the second derivative. Calculus problems often involve such methodical processes:
The exercise was a classic calculus problem where we were asked to find a second derivative. The steps took us from breaking down the given function into parts, calculating their derivatives, and applying these understandings to find the second derivative. Calculus problems often involve such methodical processes:
- Break down functions: Analyze each part of the function separately.
- Derivatives: Differentiate to find rates of change.
- Combine results: Put together the pieces for a complete solution.
Other exercises in this chapter
Problem 192
For the following exercises, find \(\frac{d^{2} y}{d x^{2}}\) for the given functions. $$y=\sin x \cos x$$
View solution Problem 192
Find \(\frac{d^{2} y}{d x^{2}}\) for the given functions. $$ y=\sin x \cos x $$
View solution Problem 193
Find \(\frac{d^{2} y}{d x^{2}}\) for the given functions. $$ y=x-\frac{1}{2} \sin x $$
View solution Problem 194
For the following exercises, find \(\frac{d^{2} y}{d x^{2}}\) for the given functions. $$y=\frac{1}{x}+\tan x$$
View solution